Not Frequent
0/10

# Knapsack DP

Authors: Nathan Chen, Michael Cao, Benjamin Qi

### Prerequisites

Problems that can be modeled as filling a limited-size container with a subset of items.

Focus Problem – read through this problem before continuing!

## Tutorial

Resources
CPHSolves "Minimizing Coins," 0/1 Knapsack

Knapsack problems generally involve filling a limited container with a subset of items, and we want to count or optimize some quantity associated with the items. Almost every time, you can think of each item as having a positive weight, and the total weight of the items we choose must not exceed the capacity of the container, which is some number. Some variations of knapsack-type problems include:

• The 0/1 Knapsack problem: Choosing a subset of items such that we maximize their total value, and their total weight does not exceed the capacity of the container
• Finding all the possible total weights that we can achieve from any subset of items, and their total weight does not exceed the capacity of the container (in the chapter of CPH linked above)
• Counting how many sequences of items will fill the container completely, meaning the total weight is exactly the capacity of the container (the order may or may not matter)

The DP solution to knapsack problems usually has the state keeping track of the capacity of the knapsack, and the transitions involve trying to add an item to the knapsack. In competitive programming, you can expect that classical knapsack problems will be given twists, disguises, and extra state information involved.

## Solution - Dice Combinations

The problem asks us how many sequences of dice rolls exist such that sum of the top faces is $N$ ($N \leq 10^6$). To keep up with the knapsack analogy, that means we have infinite numbers of items of weights $1$ through $6$, and we want to count how many sequences of items exist such that if we put items into the container while following the sequence, the container becomes completely full. Note that the order of the items matters in this problem.

For convenience, let $\texttt{dp}[x]$ be the number of sequences of dice rolls that add up to $x$. To count how many sequences add up to $N$, or in other words, to find $\texttt{dp}[N]$, let's look at the last dice roll that brings us up to a total sum of $N$.

If the last roll was a $1$, then we know there are $\texttt{dp}[N-1]$ ways to achieve sum $N$ when the last roll is $1$. If the last roll was a $2$, then we know there are $\texttt{dp}[N-2]$ ways to achieve sum $N$ when the last roll is $2$. Continue this logic for all the dice numbers up to $6$. Considering all those cases together, we have shown that

$\texttt{dp}[N] = \texttt{dp}[N-1] + \texttt{dp}[N-2] + \texttt{dp}[N-3] + \texttt{dp}[N-4] + \texttt{dp}[N-5] + \texttt{dp}[N-6].$

Apply that same logic we used for $\texttt{dp}[N]$ on a general $x$:

$\texttt{dp}[x] = \sum_{i=1}^6\texttt{dp}[x-i].$

Start with the base case that $\texttt{dp} = 1$, and then $\texttt{dp}$, $\texttt{dp}$, $\texttt{dp}$, and so on... can be calculated using the recurrence until we find $\texttt{dp}[N]$. Note in the code that we ignore $\texttt{dp}[x]$ if $x < 0$.

C++

1#include <bits/stdc++.h>2
3using namespace std;4
5typedef long long ll;6
7int main() {8    int n;9    cin >> n;10    ll dp[n+1]{}; dp = 1;

Java

1import java.util.*;2import java.io.*;3
4public class Main {5    public static void main(String[] args) throws Exception {6        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));7        int n; n = Integer.parseInt(br.readLine());8
9        long dp[] = new long[n+1];10        dp = 1;

Python

## Problems

### CSES

StatusSourceProblem NameDifficultyTagsSolution
CSESVery Easy
Show Tags

Knapsack

CSESEasy
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Knapsack

CSESEasy
Show Tags

Knapsack

CSESEasy
Show Tags

Knapsack

External Sol
CSESEasy
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Knapsack

External Sol
CSESEasy
Show Tags

Knapsack

External Sol

### USACO

StatusSourceProblem NameDifficultyTagsSolution
GoldEasy
Show Tags

DP, Knapsack

External Sol
GoldHard
Show Tags

DP, Knapsack, Binary Search

External Sol
PlatVery HardExternal Sol